Q. A particle is dropped from a height h. Another particle which is initially at a horizontal distance d from the first is simultaneously projected with a horizontal velocity u and the two particles just collide on the ground ,then:
(a) $\displaystyle d^2 = \frac{u^2 h}{2g}$
(b) $ \displaystyle d^2 = \frac{2 u^2 h}{g}$
(c) d = h
(d) g d2 = u2
Click to See Answer :
Ans: (b)